What's the question?
If you want the maximum height:
the vertex will tell you how high it went and how long it took to reach that height.
the t of the vertex is -64/(-32) = 2 seconds
h(2) = -64 + 128 + 192 = 256 ft above the ground
(the vertex was (2, 256)
when will it hit the ground?
0 = -16t^2 + 64t + 192
t^2 - 4t - 12 = 0
(t-6)(t+2) = 0
t = 6 or t = -2, the last answer makes no sense.
The height above ground of a snowball thrown from a cliff is modeled by the function h(t) = -16t2 + 64t + 192, where h is height in feet and t is time in seconds.
I d3on't know what to do with this i use to do this but know i forgot
Guide me through this please
2 answers
First you must find the vertex. The vertex is -64/(-32) = 2 seconds
h(2) = -64 + 128 + 192 = 256
V = (2/256)
To find the vertex the formula f(x) = ax^2 +bx + c must be used.
h(2) = -64 + 128 + 192 = 256
V = (2/256)
To find the vertex the formula f(x) = ax^2 +bx + c must be used.